Airfoil Vortex Sheet Models 2

Fluids – Lecture 2 Notes 1. Airfoil Vortex Sheet Models 2. Thin-Airfoil Analysis Problem Reading: Anderson 4.4, 4.7

Airfoil Vortex Sheet Models Surface Vortex Sheet Model An accurate means of representing the ﬂow about an airfoil in a uniform ﬂow is to place a vortex sheet on the airfoil surface. The total velocity V~ (x, z), which is the vector sum of the freestream velocity and the vortex-sheet velocity, can be forced parallel to the airfoil surface by suitably setting the sheet strength distribution γ(s). V γ (s) V

A panel method is normally used to numerically compute γ(s). By using a sufficient number of panels, this result can be made as accurate as needed. The main drawback of such numerical calculations is that they give limited insight into how the flow is influenced by changes in the angle of attack or the airfoil geometry. Such insight, which is important for effective aerodynamic design and engineering, is much better provided by simple approximate analytic solutions. The panel method can still be used for accuracy when it’s needed.

Single Vortex Sheet Model In order to simplify the problem sufficiently to allow analytic solution, we make the following assumptions and approximations: 1) The airfoil is assumed to be thin, with small maximum camber and thickness relative to the chord, and is assumed to operate at a small angle of attack, α ≪ 1. 2) The upper and lower vortex sheets are superimposed together into a single vortex sheet γ = γu + γℓ, which is placed on the x axis rather than on the curved mean camber line Z =(Zu + Zℓ)/2. z z z γ (x) n u Z u (x) Z(x) Z(x) x x x γ (x) γ Z l (x) γ (x) l (x) 3) The flow-tangency condition V~ · nˆ = 0 is applied on the x-axis at z = 0, rather than on the camber line at z = Z. But the normal vectorn ˆ is normal to the actual camber line shape, as shown in the figure. 4) Small-angle approximations are assumed. The freestream velocity is then written as follows.

V~∞ = V∞ (cos α)ˆı + (sin α)kˆ ≃ V∞ ˆı + αkˆ h i 1 On the x-axis where the vortex sheet lies, the sheet’s velocity w(x), which is strictly in the z-direction, is given by integrating all the contributions along the sheet.

c γ(ξ) dξ w(x) = − 0 Z 2π(x − ξ) Adding this to the freestream velocity then gives the total velocity.

c γ(ξ) dξ V~ (x, 0) = V~∞ + wkˆ ≃ V∞ˆı + V∞α − kˆ (1) 0 Z 2π(x − ξ)!

The normal unit vector is obtained from the slope of the camberline shape Z(x).

dZ ~n ~n(x) = − ˆı + kˆ , nˆ = (2) dx |~n| z n dξ γ dZ/dx V V x α w V 0 ξ x c

To force the total velocity to be parallel to the camberline, we now apply the ﬂow tangency condition V~ · nˆ = 0. Performing this dot product between (1) and (2), and removing the unnecessary factor 1/|~n| gives the fundamental equation of thin airfoil theory .

dZ c γ(ξ) dξ V∞ α − − =0 (for0

Thin-Airfoil Analysis Problem Flow tangency imposition For a given camberline shape Z(x) and angle of attack α, we now seek to determine the vortex strength distribution γ(x) such that the fundamental equation (3) is satisfied at every x location. As shown in the figure, this will result in the total velocity V~ at every x-location to be approximately parallel to the local camberline, producing a physically-correct flow about this camberline. The thinner the airfoil, the closer the camberline is to the x-axis where the flow tangency is actually imposed, and the more accurate the approximation becomes. Com- pared to typical airfoils, the height of the camberline in the figure is exaggerated severalfold for the sake of illustration. n

V V . n = 0

2 Coordinate transformation To enable solution of equation (3), it is necessary to ﬁrst perform a trigonometric substitution for the coordinate x, and the dummy variable of integration ξ.

c θο x = (1 − cos θ ) 2 o θ c ξ = (1 − cos θ) dθ 2 c dξ = sin θ dθ θ = 0 θ = π 2 dξ x 0 ξ c/2 x c As shown in the ﬁgure, θ runs from 0 at the leading edge, to π at the trailing edge. Since x and θ are interchangable, functions of x can now be treated as functions of θ. Equation (3) then becomes 1 π γ(θ) sin θ dθ dZ = V∞ α − (for 0 < θo < π) (4) 0 2π Z cos θ − cos θo dx ! where the known camberline slope dZ/dx is now considered a function of θo. This is an integral equation which must be solved for the unknown γ(θ) distribution, with the additional requirement that it satisfy the Kutta condition at the trailing edge point, γ(π)=0

Symmetric airfoil case In practice, the camberline slope dZ/dx can have any arbitrary distribution along the chord. For simplicity, we will first consider a symmetric airfoil . This has a flat camberline, with Z = 0 and dZ/dx = 0. Equation (4) then simplifies to

1 π γ(θ) sin θ dθ = V∞α (for 0 < θo < π) (5) 0 2π Z cos θ − cos θo Solution of this equation is still formidable, and is beyond scope here. Let us simply state that the solution is

1 + cos θ c − x γ(θ)= 2αV∞ or γ(x) =2αV∞ sin θ s x The shape of these distributions is shown in the ﬁgure below.

3 γ 2αV 2

0 0 π/2θ π 0.00.5 x/c 1.0

3 Note that at the trailing edge, γ = 0 as required by the Kutta condition, and that at the leading edge γ → ∞. The latter is of course not physical, although the singularity is weak (integrable), and the integrated results for cℓ and cm are in fact valid.

The load distribution pℓ − pu is obtained using the Bernoulli equation, together with the tangential velocity jump properties across the vortex sheet.

1 2 1 2 1 pℓ − pu = po − ρVℓ − po − ρVu = ρ (Vu + Vℓ)(Vu − Vℓ) = ρV∞γ 2 2 2 The lift/span on an element dξ of the sheet is

′ dL = (pℓ − pu) dξ = ρV∞γ dξ

and the total lift/span is then obtained by integrating this load distribution.

c ′ L = ρV∞ γ(ξ) dξ 0 Z The integral is also seen to be the overall circulation, making this lift result consistent with the Kutta-Joukowsky Theorem.

c ′ Γ = γ(ξ) dξ, L = ρV∞Γ 0 Z z dL'

γ dM' Vu x 0 c dξ Vl ξ

The actual integration of the loading or sheet strength is most easily performed in the trigonometric coordinate θ. By direct substitution, we have c π π Γ = γ(θ) sin θ dθ = αcV∞ (1 + cos θ) dθ = παcV∞ 0 0 2 Z Z The lift/span is then ′ 2 L = ρV∞Γ = παcρV∞ and the corresponding lift coeﬃcient is L′ cℓ = 1 2 = 2π α 2 ρV∞ c This is a very important result, showing that the lift is proportional to the angle of attack, with a lift slope of dc ℓ = 2π dα These results very closely match the results of more complex panel method calculations, as well as experimental data.

4 The pitching moment/span on the sheet element, taken about the leading edge, is obtained by weighting the lift by the moment arm ξ.

′ ′ dMLE = −ξ dL = −ρV∞γξ dξ

The overall moment/span is then

c ′ MLE = −ρV∞ γξ dξ 0 Z We can again most easily integrate this in the trigonometric coordinate.

2 π 2 π 2 ′ c 2 c 2 2 c 2 MLE = −α ρV∞ (1 + cos θ)(1 − cos θ) dθ = −α ρV∞ (1 − cos θ) dθ = −πα ρV∞ 0 0 2 Z 2 Z 4 The moment coeﬃcient about the leading edge point is

′ MLE πα cℓ cm,le = 1 2 2 = − = − 2 ρV∞ c 2 4

and the equivalent moment coeﬃcient about the standard quarter-chord point at x/c =1/4 is 1 c 4 = c le + c = 0 m,c/ m, 4 ℓ This very important result shows that a symmetric airfoil has zero moment about the quarter- chord point, for any angle of attack.